Skip to main content

Full text of "Treatise On Analysis Vol-Ii"

See other formats


spaces, and if /,- is a (complex) measure on Xf for each /, then the unique


measure v on X = J| Xf which satisfies the relation


(f(                      Wf          x)=  fr (V-Oc)d  6c)

j/^)/^) /(*) vcx15...;xj   nj-M**) M*J

is called the product of the measures /^ (1 g f ^ ), and is denoted by


Hi  ^2 " *"  Mn or (8) A*i  The existence and uniqueness of v are proved

by induction on n, putting v = (^      ^.j)  //n, and observing that if
v' also satisfies the required conditions, then we must have

for all /z e Jfl JJ Xf J, by the inductive hypothesis. This characterization of the

product of measures shows immediately that the product is associative; in
particular, we have also

We write

     /d/*! djH2    djurt   or         ' * *

instead of \fdv, and analogous notations for upper and lower integrals. We

shall leave to the reader the task of formulating and proving the results
corresponding to those established above for the case n = 2; the proofs
require nothing but simple arguments by induction on n,

The product of the Lebesgue measures on the n factors of Rw is called
Lebesgue measure on Rw.


1. Let X be a locally compact space, \i a positive measure on X, and / a real-valued
function ^0 defined on X, In the product space X x R, let D/ denote the set of points
(x, t) such that 0 <^ / ^/(x). Also let A denote Lebesgue measure on R.
(a) For/to be ^-measurable it is necessary and sufficient that D/ should be measur-
able with respect to the product measure v = p,  X. (To show that the condition is
necessary, prove that if/is ju-measurable then D/ is the union of a v-negligible set and
a denumerable family of sets of the form A x I, where A is ft-measurable and I is an
interval in R+. To show that the condition is sufficient, prove that if it is satisfied
there exists a dense subset H of R such that/""^ [a, +00]) is ^-measurable for all
a e H, by using (13.21.10).) (13.21.13) and ( that/ g is v-measurable.